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Question

A combination of secondary cells has an equivalent EMF of 2 V, internal resistance of 400 Ω. Maximum current that can be drawn is.

A
0.005 A
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B
0.0005 A
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C
0.010 A
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D
0.010 A
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Solution

The correct option is A 0.005 A
Ok. So what the question tells you is that the battery comes with a resistance of 400Ω of its own. Now we want to maximize the current we can draw from this resistor. That can only be a maximum if we connect the least possible resistance across it. That would be 0 Ω. Connecting 0 Ω across the terminals of the battery is called shorting the terminals of the battery. Now that means we have a circuit like this –

The current will simply be i=VR
i=2400=0.005A

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